3 added 375 characters in body edited Apr 8 '16 at 10:55 Chet Miller 7,42811 gold badge77 silver badges1414 bronze badges Try startingStarting with $$dG=-SdT+VdP$$ So, we have $$S=-\left(\frac{\partial G}{\partial T}\right)$$$$\left(\frac{\partial G}{\partial T}\right)_P=-S\tag{1}$$ SoBut, from the equation $$\left(\frac{\partial S}{\partial T}\right)_P=-\left(\frac{\partial^2 G}{\partial T^2}\right)_P$$$$G = H - TS$$, we also have:$$\left(\frac{\partial G}{\partial T}\right)_P=\left(\frac{\partial H}{\partial T}\right)_P-S-T\left(\frac{\partial S}{\partial T}\right)_P\tag{2}$$ If we combine Eqns. 1 and 2, we obtain:$$\left(\frac{\partial H}{\partial T}\right)_P=T\left(\frac{\partial S}{\partial T}\right)_P\tag{3}$$The left hand side of Eqn. 3 is the definition of $$C_P$$. So,$$C_P=T\left(\frac{\partial S}{\partial T}\right)_P\tag{4}$$ QED Try starting with $$dG=-SdT+VdP$$ So, $$S=-\left(\frac{\partial G}{\partial T}\right)$$ So, $$\left(\frac{\partial S}{\partial T}\right)_P=-\left(\frac{\partial^2 G}{\partial T^2}\right)_P$$ Starting with $$dG=-SdT+VdP$$, we have $$\left(\frac{\partial G}{\partial T}\right)_P=-S\tag{1}$$ But, from the equation $$G = H - TS$$, we also have:$$\left(\frac{\partial G}{\partial T}\right)_P=\left(\frac{\partial H}{\partial T}\right)_P-S-T\left(\frac{\partial S}{\partial T}\right)_P\tag{2}$$ If we combine Eqns. 1 and 2, we obtain:$$\left(\frac{\partial H}{\partial T}\right)_P=T\left(\frac{\partial S}{\partial T}\right)_P\tag{3}$$The left hand side of Eqn. 3 is the definition of $$C_P$$. So,$$C_P=T\left(\frac{\partial S}{\partial T}\right)_P\tag{4}$$ QED Post Undeleted by Chet Miller occurred Apr 8 '16 at 10:43 Post Deleted by Chet Miller occurred Apr 7 '16 at 11:41 2 added 160 characters in body edited Apr 7 '16 at 11:40 Chet Miller 7,42811 gold badge77 silver badges1414 bronze badges Try starting with $$dG=-SdT+VdP$$ So, $$S=-\left(\frac{\partial G}{\partial T}\right)$$ So, $$\left(\frac{\partial S}{\partial T}\right)_P=-\left(\frac{\partial^2 G}{\partial T^2}\right)_P$$ Try starting with $$dG=-SdT+VdP$$ Try starting with $$dG=-SdT+VdP$$ So, $$S=-\left(\frac{\partial G}{\partial T}\right)$$ So, $$\left(\frac{\partial S}{\partial T}\right)_P=-\left(\frac{\partial^2 G}{\partial T^2}\right)_P$$ 1 answered Apr 6 '16 at 19:40 Chet Miller 7,42811 gold badge77 silver badges1414 bronze badges Try starting with $$dG=-SdT+VdP$$